EJC Physics 2021 J2 H1 MYE P1 MS

©EJC 2021 8867/J2H1MYE/2021 EUNOIA JUNIOR COLLEGE JC2 Mid Year Examination 2021 General Certificate of Education Advanced Level Higher 1 PHYSICS MARK SCHEME 8867 July 2021 Paper 1 Multiple Choice Question Key Question Key Question Key 1 A 6 C 11 D 2 D 7 B 12 B 3 D 8 D 13 B 4 D 9 D 14 C 5 D 10 A 15 A 16 B 21 A 26 D 17 A 22 A 27 D 18 B 23 A 28 D 19 B 24 D 29 A 20 D 25 A 30 C Question Solution 1 Unit of v = m s─1 Unit of gλ = [ (m s─2)(m) ] ½ = [ m2 s─2 ] ½ = m s─1 2 Volume of cylinder V = A x L = π(D2/4) x L ∆𝑉𝑉 𝑉𝑉 = 2 ∆𝐷𝐷 𝐷𝐷 + ∆𝐿𝐿 𝐿𝐿 = 2 (0.02) + 0.01 = 0.05 = 5% 3 vPE = vPW + vWE where vPE is velocity of plane relative to Earth vPW is velocity of plane relative to Earth vWE is velocity of wind relative to Earth sin θ = 85/200 θ = 25.2o west of north θ vPE VWE = 85 km h─1 VPW = 200 km h─1

2 ©EJC 2021 8867/J2H1MCT/2021 4 Taking upwards as positive, uy = 3.0 m s-1 At the ground, the vertical displacement sy = - 2.4 m vy2 = uy2 + 2aysy (3.0)2 = uy2 + 2(-9.81)(-2.4) vy = -7.49 m s-1 vx = 5.0 m s-1 since there is not change in horizontal velocity v = �𝑣𝑣𝑥𝑥2 + 𝑣𝑣𝑦𝑦2 = 9.0 m s-1 5 Distance covered by car = x + (17.0 + 3.5(2)) Distance covered by lorry = x Distance covered by car - Distance covered by lorry = ½ (16)T [x + (17.0 + 3.5(2)) – x] = ½ (16)T T = 3.0 s 6 Acceleration is defined as the rate of change of velocity with time. Hence, a = (v – u)/t 7 p∆= area under net force-time graph p∆= ×area under graph cross-sectional area f f v v −−− −= ×× ××× = 3624 -1 1)( 0) (100 10 )(3.0(3.2 10 10 ) ( ) 2 1300 m 2.8 10 s 8 Area under resultant force-time graph = change in momentum Area S = mv – 0 area S v m= 9 Apply “F = ma” to the system of wagons from 2 to 6. Taking LHS as positive F = ma T ─ 5(4000) = 5(6.0 x 104) (0.15) T = 65 000 N 10 By conservation of momentum, Sum of total initial P = Sum of total final P 0 = ML(VL) + Ms(Vs) , L = large, s = small 𝑉𝑉𝐿𝐿 𝑉𝑉𝑠𝑠 = 𝑀𝑀𝑠𝑠 𝑀𝑀𝐿𝐿 = 1 3 = 0.3333 Car begins to overtake: time t = 0 x lorry car lorry Car safely overtakes: time t = T car

3 ©EJC 2021 8867/J2H1MCT/2021 11 In order for the right hinge to be in equilibrium, the forces on it are as shown. P is pulling the right hinge and R is pushing the right hinge. Hence P is in tension and R is in compression. R is in compression, hence pushing the bottom hinge. Forces by Q and S on the hinge have to be in the directions drawn for the hinge to be in equilibrium. Hence Q is in tension and S in compression. 12 Initially, W = 3T W =3kx …(1) Finally, 3W = 2T’ 3W = 2kx’ …(2) x' = 9x/2 13 Network workdone = work done against friction + gain in GPE [No gain in KE] = 580 x 500 + 1000(9.81)(87)